Hi, my question is :
Let n be a positive integer. Find all real numbers x such that $\displaystyle (x-i)^n + (x+i)^n = 0 $
I find that there is no such REAL number.
Does someone find a different answer?
Well a pretty obvious solution from inspection is that if n is even, then x = 0 is a real number solution. But I'll do the working just to see if any others lurk:
$\displaystyle (x-i)^n = -(x+i)^n $
$\displaystyle x-i = (-(x+i)^n)^{\frac{1}{n}} $
$\displaystyle x-i = (-1)^{\frac{1}{n}} (x+i) $
$\displaystyle x-i = \frac{(-1)^1}{(-1)^n} \times (x+i) $
$\displaystyle x-i = \frac{-1}{(-1)^n} \times (x+i) $
$\displaystyle x-i = \frac{-x}{(-1)^n} - \frac{i}{(-1)^n} $
Get all x terms on LHS and non-x terms on RHS
$\displaystyle x+ \frac{x}{(-1)^n} = i- \frac{i}{(-1)^n} $
$\displaystyle x(1+ \frac{1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n} $
$\displaystyle x(\frac{(-1)^n+1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n} $
$\displaystyle x = \frac{i(-1)^n-i}{(-1)^n} \times \frac{(-1)^n}{(-1)^n+1} $
$\displaystyle x = \frac{i(-1)^n-i}{(-1)^n+1} $
$\displaystyle x = \frac{i((-1)^n-1)}{(-1)^n+1} $
Hence $\displaystyle n \neq 2k+1 $ for $\displaystyle k$ any positive integer.
For $\displaystyle n = 2k$ $\displaystyle x = 0 $
Hence the equation has a real number solution for all even values of n. And that solution is $\displaystyle x = 0$. 0 is a real number.
OK this is driving me mad.
According to my CAS, it reduces to...
$\displaystyle 2\cos{\left(n\arctan{x} - \frac{n\pi}{2}\right)}(x^2 + 1)^{\frac{n}{2}}$
and gives
$\displaystyle x = -\cot{\left(\frac{k\pi}{n} - \frac{\pi}{2n}\right)}$ and $\displaystyle -1 \leq \frac{n + 2k - 1}{n} \leq 1$.